Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $2.6$ years; the standard deviation is $0.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living less than $4.1$ years.
Solution: $2.6$ $2.1$ $3.1$ $1.6$ $3.6$ $1.1$ $4.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $2.6$ years. We know the standard deviation is $0.5$ years, so one standard deviation below the mean is $2.1$ years and one standard deviation above the mean is $3.1$ years. Two standard deviations below the mean is $1.6$ years and two standard deviations above the mean is $3.6$ years. Three standard deviations below the mean is $1.1$ years and three standard deviations above the mean is $4.1$ years. We are interested in the probability of a lizard living less than $4.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lizards will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $1.1$ years and the other half $({0.15\%})$ will live longer than $4.1$ years. The probability of a particular lizard living less than $4.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.